Problem 80 - Chemistry: The Central Science

Question

Write a balanced equation for each of the following reactions: (a) Diborane reacts with water to form boric acid and molecular hydrogen. (b) Upon heating, boric acid undergoes a condensation reaction to form tetraboric acid. (c) Boron oxide dissolves in water to give a solution of boric acid.

Step-by-Step Solution

Answer

a) Balancing the equation for the reaction of diborane with water: \(B_2H_6 + 6H_2O \rightarrow 2H_3BO_3 + 3H_2\) b) Balancing the equation for the condensation reaction of boric acid: \(4H_3BO_3 \rightarrow H_2B_4O_7 + 5H_2O\) c) Balancing the equation for the reaction of boron oxide with water: \(B_2O_3 + 3H_2O \rightarrow 2H_3BO_3\)

Step 1: Identify the compounds

Diborane: B2H6, Water: H2O, Boric acid: H3BO3, Molecular hydrogen: H2

Step 2: Write the unbalanced equation

B2H6 + H2O -> H3BO3 + H2

Step 3: Balance the equation

To balance the equation, we adjust the stoichiometric coefficients: B2H6 + 6H2O -> 2H3BO3 + 3H2 b) Upon heating, boric acid undergoes a condensation reaction to form tetraboric acid.

Step 4: Identify the compounds

Boric acid: H3BO3, Tetraboric acid: H2B4O7

Step 5: Write the unbalanced equation

H3BO3 -> H2B4O7

Step 6: Balance the equation

To balance the equation, we adjust the stoichiometric coefficients: 4H3BO3 -> H2B4O7 + 5H2O c) Boron oxide dissolves in water to give a solution of boric acid.

Step 7: Identify the compounds

Boron oxide: B2O3, Water: H2O, Boric acid: H3BO3

Step 8: Write the unbalanced equation

B2O3 + H2O -> H3BO3

Step 9: Balance the equation

To balance the equation, we adjust the stoichiometric coefficients: B2O3 + 3H2O -> 2H3BO3